To determine which compound is white in color, we need to examine the properties of each option. Let's evaluate them one by one:
From the above analysis, we can conclude that among the listed options, lead sulphate is distinctly known for its white coloration.
Therefore, the correct answer is lead sulphate.
Lead sulphate is white in color.
Ammonium sulphide is soluble.
Lead iodide is bright yellow.
Ammonium arsenomolybdate is yellow.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
| (a) \([Cr(H_2O)_6]^{+3}\) | (i) \(t^2_{2g}eg^0\) |
| (b) \([Fe(H_2O)_6]^{+3}\) | (ii) \(t^3_{2g}eg^0\) |
| \((c) [Ni(H_2O)_6]^{+2}\) | (iii) \(t^3_{2g}eg^2\) |
| (d) \([V(H_2O)_6]^{+3}\) | (iv) \(t^6_{2g}eg^2\) |
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,